Past exam of the options II course of the University of Oxford / 2012 / E / E4

The quotient topology is defined byPreimages commute with arbitrary unions and finite intersections and take the empty and whole sets to the corresponding sets, so this definition indeed gives a topology.

A map is continuous exactly when the preimage of every open set is open. By the definition in part (i), every open $U\subseteq X/\mathcal{R}$ has open preimage $p^{-1}(U)$. Thus $p$ is continuous, directly from the defining property of the quotient topology.

If $g$ is continuous, so is the composition $g\circ p$. Conversely, assume the composition is continuous. For every open set $V\subseteq Y$,is open in $X$. The quotient definition then says that $g^{-1}(V)$ is open in $X/\mathcal{R}$. Hence $g$ is continuous andThis is the universal property of the quotient topology.

Let two distinct quotient points correspond to two equivalence classes. By hypothesis those classes lie in disjoint open saturated sets $U,V\subseteq X$. Saturation gives $p^{-1}(p(U))=U$, so the quotient definition makes $p(U)$ open, and likewise $p(V)$.

The image sets contain the two quotient points and are disjoint. Indeed, a class meeting both $U$ and $V$ would be contained in each because both sets are saturated, contradicting their disjointness. Thus the quotient is Hausdorff.

Define $\tilde{f}:X/\mathcal{F}\to Y$ by $\tilde{f}([x])=f(x)$. Its definition is independent of the representative because the classes are precisely the fibres of $f$. It is onto since $f$ is onto, and injective because two points with equal $f$-values lie in the same fibre.

The relation $\tilde{f}\circ p=f$ and part (iii) imply continuity of $\tilde{f}$. Since $X$ is compact and $p$ is continuous, $X/\mathcal{F}$ is compact. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Therefore

Every iterate $g^n(x,y)=(x+n,(-1{)}^{n}y)$ is an isometry of the strip with its Euclidean subspace metric. Let $a,b$ lie in distinct orbits. Since the first-coordinate difference in $a-g^{n}b$ grows without bound as $|n|\to \infty$, the positive distances $d(a,g^{n}b)$ have a positive minimum $\delta$. Only finitely many indices can give a distance below any fixed bound, and none gives zero.

Choose $0<\epsilon <\delta /3$. Saturate the relative open balls by settingThese are open saturated neighbourhoods of the two orbits. Any two centres $g^{n}a,g^{m}b$ are at least $\delta$ apart, since their distance equals $d(a,g^{m-n}b)$. Their radius-$\epsilon$ balls therefore cannot intersect. Hence $U\cap V=\varnothing$, and part (a)(iv), an instance of positive orbit separation implies a Hausdorff quotient, proves

Let $D=[0,1]\times [-1,1]$ and $q=p\circ i:D\to S/{\mathcal{R}}^{\prime }$. This is continuous. It is onto: given $(x,y)\in S$, choose an integer shift bringing $x$ into $[0,1]$, with the corresponding possible sign change in $y$; the resulting representative lies in $D$.

Two representatives in $D$ are in the same orbit only if their first coordinates differ by an integer. An interior point can therefore be equivalent only to itself. Distinct boundary representatives have first coordinates zero and one, and the iterate $g$ identifies exactly $(0,y)$ with $(1,-y)$. Thus the fibres of $q$ are precisely the equivalence classes defining the Möbius band $M$.

The rectangle $D$ is compact, and the target is Hausdorff by part (i). Applying part (a)(v) to $q$ therefore yieldsThe compactness argument supplies continuity of the inverse without requiring a discontinuous choice of one representative per orbit.